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Thermo – the 2 nd Law Two factors effect any process such as a chemical reaction. 1.the difference in any energy term, ΔE, ΔH or ΔS, between the reactants.

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Presentation on theme: "Thermo – the 2 nd Law Two factors effect any process such as a chemical reaction. 1.the difference in any energy term, ΔE, ΔH or ΔS, between the reactants."— Presentation transcript:

1 Thermo – the 2 nd Law Two factors effect any process such as a chemical reaction. 1.the difference in any energy term, ΔE, ΔH or ΔS, between the reactants and products. That is between the final and initial states. 2.The pathway the process follows.

2 Thermo – the 2 nd Law Thermodynamics is the study of the energy of the initial and final states. Kinetics is the study of the pathway between these states.

3 Thermo – the 2 nd Law A Conundrum or two A ball spontaneously rolls downhill but never up. Water at a temperature below 0 o C freezes spontaneously, but above 0 o C ice melts. And the freezing of water is exothermic, while the melting is endothermic. So why are these spontaneous?

4 Thermo – the 2 nd Law There are obviously two different thermodynamic properties at play here. These are Enthalpy and an new one, Entropy. Entropy is a measure of the positional freedom of a set of particles.

5 Thermo – the 2 nd Law In any spontaneous process the entropy of the universe increases. Or You not only can’t win, you can’t break even.

6 Thermo – the 2 nd Law ΔS, the change in entropy, is a measure of the change in positional freedom of a set of particles. E.g. B&G spreads salt on the sidewalks. How does the entropy of the salt change as it dissolves? As the ice melts? As the water evaporates? As the salt recrystalizes?

7 Thermo – the 2 nd Law Gibbs Free Energy The maximum energy available for any thermodynamic process, G. G = H – TS & ΔG = ΔH – TΔS Note how ΔG changes as ΔH, goes from 0, as ΔS goes from 0 and as T changes.

8 Thermodynamics the Isothermal expansion of an ideal gas ΔT = 0, therefore ΔE = 0 and –q = w (or q = -w). If we let V change in infinitesimal steps we find, w = -  P external dV from V i to V f, or w = -nRT(lnV f -V i ) = -nRT*ln(V f /V i ) = -q

9 Thermodynamics the Adiabatic expansion of an ideal gas No heat may flow into or out of the system. q=0, ΔE = w = -PΔV  = C P and C P = 5/2R while C V =3/2R C V (for a monatomic gas) P i V i  =P f V f  we can use the ideal gas equation to find ΔT and, since ΔE = w = nC V ΔT, we can find the work.

10 Thermodynamics the Adiabatic expansion of an ideal gas IsothermalP i V i = P f V f or PV is constant AdiabaticP i V i  = P f V f  or PV  is constant


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